solution1 - EL630 Homework1 Solutions 1 a b c d A ={1 on...

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Unformatted text preview: EL630: Homework1 Solutions 1) a) b) c) d) A = {1 on first roll, 5 on second roll}, B = {4 on first roll, 5 on second roll} A = {even on first roll}, B = {odd on second roll} A = φ , B = any event A = B = {even on first roll} 2) Π( Α Β) = Π( ΑΒ) Π( Α) 5 = =, Π( Β) Π( Β) 8 Π( Β Α) = Π( ΑΒ) Π( Α) = =1 Π( Α) Π( Α) 3) A = { A, C}U{ A, D} where the events { A, C} and { A, D} are mutually exclusive. 1 1 Therefore, Π( Α) = Π( Α, Χ) + Π( Α, ∆ ) = . Similarly, Π(Χ) = Π( Α, Χ) + Π( Β, Χ) = . 2 2 1 Since P( A) P (C ) = ≠ P ( A, C ) , A and C are not independent. 4 4) Let A = {both are boys}, B = {the older is a boy}, C = {the younger is a boy}, D = {at least one is a boy}. Assume that the sexes of the two children are independent.Then Π( Α Β) = Π( Α ∆ ) = Π( ΑΒ) Π( Α) 1 / 4 1 = = = Π( Β) Π( Β) 1 / 2 2 Π( Α∆ ) Π( Α) 1/ 4 1 = = = Π( ∆ ) Π( ∆ ) 1 − 1 / 4 3 5) Let A = {at least one defective bulb}, B = {at least two defective bulbs}, C = {no defective bulbs}, D = {exactly one defective bulb}, E = {at most one defective bulb}. a) Π(Χ) = 1 − Π( Α) = 0.9 . b) Π( ∆ ) = Π( ΑΒ ) = Π( Α) − Π( Β) = 0.06 . The second equality follows from the fact that B ⊂ A , so P( A) = P ( B ) + P( AB ) . c) E = C U D , and C I D = φ , so Π( Ε ) = Π(Χ) + Π( ∆ ) = 0.96 . 6) Let M = {Person is a man}, P( M B) = P( B M ) P( M ) P( B) = W = {person is a woman}, B = {person is blonde}. P( B M ) P( M ) P( B M ) P ( M ) + P( B W ) P (W ) = (0.3)(0.6) = 0.5294 (0.3)(0.6) + (0.4)(0.4) ...
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This note was uploaded on 04/24/2011 for the course ECE 6303 taught by Professor Voltz during the Spring '11 term at NYU Poly.

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solution1 - EL630 Homework1 Solutions 1 a b c d A ={1 on...

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